Journal of Inequalities in Pure and
Applied Mathematics
ON A F. QI INTEGRAL INEQUALITY
ALFRED WITKOWSKI
volume 6, issue 2, article 36,
2005.
Mielczarskiego 4/29
85-796 Bydgoszcz, Poland
Received 16 August, 2004;
accepted 06 February, 2005.
EMail: alfred.witkowski@atosorigin.com
Communicated by: F. Qi
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
152-04
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Abstract
Necessary and sufficient conditions under which the Qi integral inequality
Z
b
f t (x) dx ≥
Z
a
b
t−1
f(x) dx
a
or its reverse hold for all t ≥ 1 are given.
2000 Mathematics Subject Classification: 26D15.
Key words: Convexity, Qi type integral inequality.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Feng Qi Integral Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Reversed Feng Qi Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4
Final Remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
References
On a F. Qi Integral Inequality
Alfred Witkowski
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1.
Introduction
In [5] Feng Qi formulated the following problem: Characterize a positive function f such that the inequality
Z b
t−1
Z b
t
(1.1)
f (x) dx ≥
f (x) dx
a
a
holds for t > 1.
In [1, 2, 3, 4] and the references therein, several sufficient conditions and
generalizations are given. In all the cited papers the authors look for the solution
of the Qi inequality with restricted t. This paper is another contribution to this
subject. We shall try to establish conditions under which the inequality holds
for all t > 1.
Let (X, µ) be a finite measure space and f be a positive measurable function.
Define for t ∈ R
Z
Z
t
(1.2)
H(t) = H(t, f ) = ln f dµ − (t − 1) ln
f dµ .
On a F. Qi Integral Inequality
Alfred Witkowski
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It is clear that inequality (1.1) is equivalent to H(t) ≥ 0 for t > 1. We will say
that for the function f the Qi Inequality (QI) holds if H(t, f ) is nonnegative for
all t ≥ 1. We will also say that for the function f the Reverse Qi Inequality
(RQI) holds if H(t, f ) is non-positive for all t ≥ 1.
By the Cauchy-Schwarz integral inequality, we have for p, q ∈ R
Z
2 Z
Z
p+q
p
2
(1.3)
f
dµ ≤ f dµ f q dµ
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which means that the function H(t) is convex, that is
t1 + t2
H(t1 ) + H(t2 )
(1.4)
H
≤
2
2
holds for t1 , t2 ∈ R, so its derivative
R t
Z
f ln f dµ
0
R
(1.5)
H (t) =
− ln
f dµ
f t dµ
On a F. Qi Integral Inequality
is increasing in t ∈ R.
Let
Alfred Witkowski
M = ess supx∈X f (x) and µM = µ({x : f (x) = M }).
The following lemmas will be useful.
Note that from now on we will use the convention that ln ∞ = ∞ and ln 0 =
−∞.
Lemma 1.1. The following formula holds:
(1.6)
H(t)
M
= ln R
.
t→∞
t
f dµ
lim
Proof. For ε > 0 let mε = µ(x : f (x) > M − ε). Then
Z
t
(M − ε) mε ≤ f t dµ ≤ M t µ(X),
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so
Z
(1.7)
t ln(M − ε) + ln mε ≤ H(t) + (t − 1) ln
f dµ
≤ t ln M + ln µ(X).
Dividing by t on both sides of (1.7) yields
M −ε
H(t)
M
H(t)
ln R
≤ lim inf
≤ lim sup
≤ ln R
.
t→∞
t
t
f dµ
f dµ
t→∞
In case M = ∞, M − ε stands for an arbitrary large number. This completes
the proof.
Lemma 1.2. If M < ∞ then
Z
M
= ln µM f dµ .
(1.8)
lim H(t) − t ln R
t→∞
f dµ
Proof. Direct computation yields
Z t
Z
M
f
lim H(t) − t ln R
(1.9)
= lim ln
dµ + ln f dµ
t→∞
t→∞
M
f dµ
Z
= ln µM f dµ
as (f /M )t tends monotonically to the characteristic function of {x : f (x) =
M }.
On a F. Qi Integral Inequality
Alfred Witkowski
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2.
Feng Qi Integral Inequality
In this section we consider the problem: Characterize positive functions f that
satisfy (QI).
Theorem 2.1. A constant function M satisfies (QI) if and only if µ(X) ≤ 1 and
M ≥ 1/µ(X).
Proof. H(t) ≥ 0 is equivalent to M ≥ µ(X)t−2 . This can be valid for all t > 1
only if the conditions of the theorem are fulfilled.
From now on we assume that f is not constant, in which case the function
H is strictly convex.
R It is clear that the necessary condition for (QI) is H(1) ≥ 0 or equivalently
f dµ ≥ 1.
On a F. Qi Integral Inequality
Alfred Witkowski
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Theorem 2.2.
(a) If H(1) = 0 then (QI) holds if and only if H 0 (1) ≥ 0.
(b) If H(1) > 0 then
0
(b1) if H (1) ≥ 0 then (QI) holds;
R
(b2) if H 0 (1) < 0 and M < f dµ then (QI) fails for large t;
R
(b3) if H 0 (1) < 0 and M = f dµ then (QI) holds if and only if µM M ≥
1;
R
(b4) if H 0 (1) < 0 and M > f dµ then there exists an unique point t0
such that H 0 (t0 ) = 0 and (QI) holds if and only if H(t0 ) ≥ 0.
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Proof. (a) and (b1) follow immediately from convexity of H.
From Lemma 1.1 we see that H becomes negative for large t, which proves
(b2).
(b3) follows from Lemma 1.2 and from the fact that being convex the graph
of H lies above its horizontal asymptote.
Finally (b4) follows from the fact that H 0 is strictly increasing and H 0 (t0 ) =
0 for some t0 , then H attains its minimum at t0 . Observe that in this case H
may be infinite for some finite t∞ and consequently for all t > t∞ .
On a F. Qi Integral Inequality
From the above theorem we obtain the following, surprising
Alfred Witkowski
Corollary 2.3. If µ(X) < 1 then (QI) holds if and only if H(1) ≥ 0.
Proof. We will show that if µ(X) < 1 then H 0 (1) ≥ 0 for all f , so the condition
(b1) is satisfied.
Applying the integral Jensen Inequality to the convex function x ln x we obtain
Z
Z
Z
1
1
1
f ln f dµ ≥
f dµ ln
f dµ
µ(X)
µ(X)
µ(X)
Z
Z
1
≥
f dµ ln
f dµ
µ(X)
which is equivalent to H 0 (1) ≥ 0.
In the case (b4), solving the equation H 0 (t) = 0 may not be an easy task, but
the following corollaries may be helpful:
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Corollary 2.4. Let
R
tL =
R
R
f ln f dµ − f dµ ln f dµ
R
.
f ln f dµ
If H 0 (tL ) ≥ 0 then (QI) holds.
Proof. tL is the point where the supporting line drawn at t = 1 meets the OXaxis. The graph of H lies above it. In particular H(tL ) ≥ 0. As H 0 (t) is
nonnegative for t ≥ tL the proof is completed.
Corollary 2.5. If 0 < µM , M < ∞ let
R
ln(µM f dµ)
R
tR = −
.
ln(M/ f dµ)
0
If H (tR ) ≤ 0 or tR ≤ tL then (QI) holds.
Proof. tR is the point where the supporting line drawn at ∞ (it exists by Lemma
1.2) meets the OX-axis. If tR ≤ tL the two supporting lines meet above the OXaxis.
If H 0 (tR ) ≤ 0 we use an argument similar to that in the proof of the previous
corollary.
On a F. Qi Integral Inequality
Alfred Witkowski
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3.
Reversed Feng Qi Inequality
In this section we give sufficient and necessary conditions for the reversed problem: Characterize positive functions f that satisfy (RQI).
Theorem 3.1. A constant function satisfies (RQI) if and only if µ(X) ≥ 1 and
M ≤ 1/µ(X).
The proof is similar to that of Theorem 2.1.
Theorem 3.2.
R For a non constant function f (RQI) holds if and only if H(1) ≤
0 and M ≤ f dµ.
R
Proof. As µM M < f dµ = exp(H(1)) ≤ 1 it follows from Lemma 1.1 and
1.2 that H is negative for large t. Being convex and non-positive at t = 1, it
must be decreasing.
R
On the other hand if M > f dµ then H is positive for large t by Lemma
1.1.
On a F. Qi Integral Inequality
Alfred Witkowski
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4.
Final Remark
Finally we prove the following
Theorem 4.1. For every positive function f there exists a constant c > 0 such
that cf satisfies (QI) or (RQI).
Proof. One can easily see that
H(t, cf ) = ln c + H(t, f ),
so cf satisfies (QI) for certain c if and only if H(t, f ) is bounded from below.
Similarly cf satisfies (RQI) only if H(t, f ) is bounded from above.
It follows immediately from Lemma 1.1 and
R Lemma 1.2 that
R the function H(t)
is bounded from below if and only if M > f dµ
or
M
=
f dµ and µM > 0
R
and is bounded from above if and only if M ≤ f dµ.
This completes the proof of our theorem.
R
Note that in case M = f dµ and µM > 0 we can find constants c1 and c2
such that (QI) holds for c1 f and (RQI) holds for c2 f .
On a F. Qi Integral Inequality
Alfred Witkowski
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References
[1] L. BOUGOFFA, Notes on Qi type integral inequalities, J. Inequal. Pure
Appl. Math., 4(4) (2003), Art. 77. [ONLINE: http://jipam.vu.edu.
au/article.php?sid=318].
[2] S. MAZOUZI AND F. QI, On an open problem regarding an integral inequality, J. Inequal. Pure Appl. Math., 4(2) (2003), Art. 31. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=269].
On a F. Qi Integral Inequality
[3] J. PECARIĆ AND T. PEJKOVIĆ, Note on Feng Qi’s integral inequality,
J. Inequal. Pure Appl. Math., 5(3) (2004), Art. 51. [ONLINE: http://
jipam.vu.edu.au/article.php?sid=418].
[4] T.K. POGANY, On an open problem of F. Qi, J. Inequal. Pure Appl.
Math., 3(4) (2002), Art. 54. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=206].
[5] F. QI, Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2)
(2000), Art. 19. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=113]. RGMIA Res. Rep. Coll., 2(7) (1999), Art. 9, 1039–1042.
[ONLINE: http://rgmia.vu.edu.au/v2n7.html].
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